leetcode801 Minimum Swaps To Make Sequences Increasing （动态规划）

leetcode801 Minimum Swaps To Make Sequences Increasing （动态规划）

Description

We have two integer sequences

A

and

B

of the same non-zero length.

We are allowed to swap elements

A[i]

and

B[i]

. Note that both elements are in the same index position in their respective sequences.

At the end of some number of swaps,

A

and

B

are both strictly increasing. (A sequence is strictly increasing if and only if

A[0] < A[1] < A[2] < ... < A[A.length - 1]

.)

Given A and B, return the minimum number of swaps to make both sequences strictly increasing. It is guaranteed that the given input always makes it possible.

Example:
Input: A = [1,3,5,4], B = [1,2,3,7]
Output: 1
Explanation:
Swap A[3] and B[3].  Then the sequences are:
A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
which are both strictly increasing.

Note:

+ A, B are arrays with the same length, and that length will be in the range [1, 1000].

+ A[i], B[i] are integer values in the range [0, 2000].

代码

参考

leetcode801. Minimum Swaps To Make Sequences Increasing(python)

class Solution:
def minSwap(self, A, B):
dp = [[len(A),len(A)] for i in range(len(A))]
# dp[i][0] 到第i个位置严格递增且不需要交换第i个数 需要交换的次数
# dp[i][1] 到第i个位置严格递增且需要交换第i个数 需要交换的次数
dp[0][0] = 0;
dp[0][1] = 1;
for i in range(1, len(A)):
if A[i] > A[i - 1] and B[i] > B[i - 1]:
dp[i][0] = dp[i - 1][0]
dp[i][1] = dp[i - 1][1] + 1

if A[i] > B[i - 1] and B[i] > A[i - 1]:
dp[i][0] = min(dp[i][0], dp[i - 1][1])
dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1)

return min(dp[len(A) - 1][0], dp[len(A) - 1][1])